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For the family of equations ∣x−4∣=εx+6|x - 4| = \varepsilon x + 6, determine the exact real solution xx in terms of ε\varepsilon for sufficiently small positive ε\varepsilon, and evaluate the asymptotic limit lim⁡ε→0+x(ε)\lim_{\varepsilon \to 0^+} x(\varepsilon) for the positive branch.

Answer and steps

Final answer

x(ε)=101−εx(\varepsilon) = \frac{10}{1 - \varepsilon}, with lim⁡ε→0+x(ε)=10\lim_{\varepsilon \to 0^+} x(\varepsilon) = 10

Steps

01

Set up the equation for the positive branch

To solve the equation ∣x−4∣=εx+6|x - 4| = \varepsilon x + 6 for the positive branch, we use the piecewise definition of the absolute value function: ∣x−4∣=x−4when x≥4|x - 4| = x - 4 \quad \text{when } x \ge 4

For the branch where x−4≥0x - 4 \ge 0 (or x≥4x \ge 4), the equation simplifies to:

x−4=εx+6x - 4 = \varepsilon x + 6

02

Solve for xx in terms of ε\varepsilon

Rearrange the terms to group all terms containing xx on one side and constant terms on the other side:

x−εx=6+4x - \varepsilon x = 6 + 4

x(1−ε)=10x(1 - \varepsilon) = 10

For sufficiently small positive ε\varepsilon (specifically, 0<ε<10 < \varepsilon < 1), the coefficient 1−ε≠01 - \varepsilon \neq 0. Dividing both sides by 1−ε1 - \varepsilon gives:

x(ε)=101−εx(\varepsilon) = \frac{10}{1 - \varepsilon}

03

Verify the domain condition

We must verify that this solution satisfies the condition x≥4x \ge 4: For any ε∈(0,1)\varepsilon \in (0, 1), we have 0<1−ε<10 < 1 - \varepsilon < 1, which implies:

x(ε)=101−ε>10>4x(\varepsilon) = \frac{10}{1 - \varepsilon} > 10 > 4

Since x>4x > 4, the expression inside the absolute value is indeed positive, confirming that x(ε)=101−εx(\varepsilon) = \frac{10}{1 - \varepsilon} is an exact valid solution.

04

Evaluate the limit as ε→0+\varepsilon \to 0^+

Taking the one-sided limit as ε\varepsilon approaches 00 from the right:

lim⁡ε→0+x(ε)=lim⁡ε→0+101−ε=101−0=10\lim_{\varepsilon \to 0^+} x(\varepsilon) = \lim_{\varepsilon \to 0^+} \frac{10}{1 - \varepsilon} = \frac{10}{1 - 0} = 10

Geometrically, this corresponds to the right-hand intersection point of the V-shaped graph y=∣x−4∣y = |x - 4| and the rotating line y=εx+6y = \varepsilon x + 6 tending toward the horizontal line y=6y = 6 at (10,6)(10, 6).

Educational illustration for this question

Typical errors

OverviewACEpath Teacher's Diagnosis

01 The student forgets to negate the expression inside the absolute value when x < 4, instead only solving the single case x - 4 = εx + 6.

Sample Student Work

Reviewed by ACEpath Teacher
Step-by-Step Review
Walk Through It with Your Teacher
Correct
  1. Step 1

    x - 4 = \varepsilon x + 6   ⟹  \implies x - \varepsilon x = 10   ⟹  \implies x(1 - \varepsilon) = 10   ⟹  \implies x = 101−ε\frac{10}{1 - \varepsilon}

  2. Step 2

    x(\varepsilon) = 101−ε\frac{10}{1 - \varepsilon}   ⟹  \implies $\lim_{\varepsilon →\to 0^+}$ \frac{10}{1 - \varepsilon} = 10_

WHAT YOU DID WELL
Compliments from Your Teacher

You correctly identified the algebraic equation for the positive branch by removing the absolute value bars as x−4x - 4.

You factored the variable terms cleanly to isolate x(ε)=101−εx(\varepsilon) = \frac{10}{1 - \varepsilon}.

You evaluated the limit as ε→0+\varepsilon \to 0^+ accurately.

THE ERROR
Where Your ACEpath Teacher Spotted the Mistake
none

None. Your written steps and final answers are completely correct.

WHY IT HAPPENED
ACEpath Teacher's Diagnosis

None. You demonstrate a clear understanding of setting up branch equations and evaluating one-sided limits.

REASONING PATH
Your ACEpath teacher’s suggested approach
Branch Domain Verification

  • Whenever you solve an absolute value equation by opening a branch (here, x−4≥0x - 4 \ge 0, so x≥4x \ge 4), always verify that the resulting parametric solution satisfies that condition.

  • Check that for any small ε∈(0,1)\varepsilon \in (0, 1), x(ε)=101−ε>10>4x(\varepsilon) = \frac{10}{1 - \varepsilon} > 10 > 4, confirming the solution legitimately lies on the positive branch.

CORRECTED PATH
ACEpath Teacher's Demonstration

  1. Verify the branch domain: since 0<ε<10 < \varepsilon < 1, we have 1−ε<11 - \varepsilon < 1, which gives x(ε)=101−ε>10≥4x(\varepsilon) = \frac{10}{1 - \varepsilon} > 10 \ge 4.

  2. The positive branch solution is fully valid and its limit as ε→0+\varepsilon \to 0^+ is 1010.

Final answer

x(ε)=101−εx(\varepsilon) = \frac{10}{1 - \varepsilon}, with lim⁡ε→0+x(ε)=10\lim_{\varepsilon \to 0^+} x(\varepsilon) = 10

NEXT STEPS
Final Takeaways from Your Teacher

  • Include a brief domain check (e.g., verifying x(ε)>4x(\varepsilon) > 4) in your written work to make your solutions fully rigorous.

  • Practice solving for the negative branch (where x<4x < 4) for the same family of equations to see how both intersection points behave as ε→0+\varepsilon \to 0^+.

OverviewACEpath Teacher's Diagnosis

02 The student makes a sign error when expanding the negative branch as -(x - 4) = -x - 4, leading to an incorrect second solution.

Sample Student Work

Reviewed by ACEpath Teacher
Step-by-Step Review
Walk Through It with Your Teacher
Partially correct
  1. Step 1

    x - 4 = \varepsilon x + 6   ⟹  \implies (1 - \varepsilon)x = 10   ⟹  \implies x = 101−ε\frac{10}{1 - \varepsilon}

  2. Step 2

    -x - 4 = \varepsilon x + 6   ⟹  \implies -(1 + \varepsilon)x = 10   ⟹  \implies x = −101+ε\frac{-10}{1 + \varepsilon}

    Correct

    -(x - 4) = \varepsilon x + 6   ⟹  \implies -x + 4 = \varepsilon x + 6   ⟹  \implies x(1 + \varepsilon) = -2   ⟹  \implies x = -21+ε\frac{2}{1 + \varepsilon}

    When expanding ∣x−4∣|x - 4| for x<4x < 4, the absolute value yields −(x−4)=−x+4-(x - 4) = -x + 4. You dropped the sign distribution on the constant, writing −x−4-x - 4, which corrupted this branch's solution.

  3. Step 3

    Positive branch is \text{Positive branch is } x(\varepsilon) = 101−ε\frac{10}{1 - \varepsilon}   ⟹  \implies $\lim_{\varepsilon →\to 0^+}$ \frac{10}{1 - \varepsilon} = 10_

WHAT YOU DID WELL
Compliments from Your Teacher

You correctly split the absolute value equation into cases.

You solved the linear equation for the positive branch with complete algebraic accuracy.

You correctly evaluated the limit as ε→0+\varepsilon \to 0^+.

THE ERROR
Where Your ACEpath Teacher Spotted the Mistake
Sign Distribution in Absolute Value Expansion

You wrote −x−4-x - 4 instead of −(x−4)=−x+4-(x - 4) = -x + 4 when removing the absolute value bars in the negative case.

WHY IT HAPPENED
ACEpath Teacher's Diagnosis

Failing to enclose the inner expression in parentheses when applying the negative branch definition ∣u∣=−u|u| = -u for u<0u < 0, leading to a missing sign change on the constant term.

REASONING PATH
Your ACEpath teacher’s suggested approach
Parentheses Around Negated Binomials

  • When replacing ∣A−B∣|A - B| with its negative branch, write −(A−B)=−A+B-(A - B) = -A + B.

  • Always wrap the entire expression inside absolute value bars in parentheses before applying the leading negative sign.

CORRECTED PATH
ACEpath Teacher's Demonstration

  1. Write the negative branch with explicit parentheses: −(x−4)=εx+6-(x - 4) = \varepsilon x + 6.

  2. Distribute the negative sign: −x+4=εx+6-x + 4 = \varepsilon x + 6.

  3. Rearrange terms: −x−εx=6−4  ⟹  −(1+ε)x=2-x - \varepsilon x = 6 - 4 \implies -(1 + \varepsilon)x = 2.

  4. Solve for xx: x=−21+εx = -\frac{2}{1 + \varepsilon}.

Final answer

x(\varepsilon) = 101−ε\frac{10}{1 - \varepsilon}  (positive branch)\text{ (positive branch)},   x(\varepsilon) = -21+ε\frac{2}{1 + \varepsilon}  (negative branch)\text{ (negative branch)},   $\lim_{\varepsilon →\to 0^+}$ x(\varepsilon) = 10_

NEXT STEPS
Final Takeaways from Your Teacher

  • Always write parentheses immediately when removing absolute value bars on the negative branch, such as −(x−4)-(x - 4).

  • Verify that both terms inside the absolute value change sign: xx becomes −x-x and −4-4 becomes +4+4.

OverviewACEpath Teacher's Diagnosis

03 The student incorrectly identifies the positive branch as x = -2/(1 + ε) by mistaking the branch with the + sign in the denominator for the positive solution.

Sample Student Work

Reviewed by ACEpath Teacher
Step-by-Step Review
Walk Through It with Your Teacher
Incorrect
  1. Step 1

    x−4=εx+6  ⟹  x=101−εx - 4 = \varepsilon x + 6 \implies x = \frac{10}{1 - \varepsilon}

  2. Step 2

    −(x−4)=εx+6  ⟹  −x+4=εx+6  ⟹  x(1+ε)=−2  ⟹  x=−21+ε-(x - 4) = \varepsilon x + 6 \implies -x + 4 = \varepsilon x + 6 \implies x(1 + \varepsilon) = -2 \implies x = \frac{-2}{1 + \varepsilon}

  3. Step 3

    Positive branch has + in denominator: x(ε)=−21+εx(\varepsilon) = \frac{-2}{1 + \varepsilon}

    Correct

    Positive branch means x−4≥0x - 4 \ge 0 (so ∣x−4∣=+(x−4)|x - 4| = +(x - 4)), which gives x(ε)=101−εx(\varepsilon) = \frac{10}{1 - \varepsilon}.

    The term 'positive branch' of an absolute value equation refers to taking the positive expression inside the absolute value, ∣u∣=+u|u| = +u when u≥0u \ge 0, not to whether a plus sign appears in the resulting denominator.

  4. Step 4

    lim⁡ε→0+−21+ε=−2\lim_{\varepsilon \to 0^+} \frac{-2}{1 + \varepsilon} = -2

    Correct

    lim⁡ε→0+101−ε=10\lim_{\varepsilon \to 0^+} \frac{10}{1 - \varepsilon} = 10

    Because the branch was misidentified in the previous step, your limit calculation evaluated the wrong branch.

WHAT YOU DID WELL
Compliments from Your Teacher

You correctly split the absolute value equation into its two linear cases.

You rearranged terms and factored out xx accurately in both equations.

You evaluated the algebraic limit of your chosen expression correctly.

THE ERROR
Where Your ACEpath Teacher Spotted the Mistake
Misinterpreted Branch Definition

You associated the phrase 'positive branch' with the '+' operator in the denominator (1+ε)(1 + \varepsilon) rather than the definition of the positive branch of the absolute value function.

WHY IT HAPPENED
ACEpath Teacher's Diagnosis

Superficial sign association: confusing the sign of an algebraic operation in an expression with the sign condition of the absolute value function's argument (x−4≥0x - 4 \ge 0).

REASONING PATH
Your ACEpath teacher’s suggested approach
Define the branches by domain

  • For ∣u∣|u|, the positive branch is u≥0u \ge 0 where ∣u∣=u|u| = u; the negative branch is u<0u < 0 where ∣u∣=−u|u| = -u.

  • Always label your two cases clearly as Case 1 (u≥0u \ge 0) and Case 2 (u<0u < 0) before solving.

Check solution validity

  • A branch solution must satisfy the domain condition of that branch.

  • Test whether 101−ε≥4\frac{10}{1 - \varepsilon} \ge 4 or −21+ε<4\frac{-2}{1 + \varepsilon} < 4 to verify which solution belongs to the x≥4x \ge 4 branch.

CORRECTED PATH
ACEpath Teacher's Demonstration

  1. Identify the positive branch of ∣x−4∣|x - 4| as the case where x−4≥0x - 4 \ge 0, which yields ∣x−4∣=x−4|x - 4| = x - 4.

  2. Select the corresponding solution from Step 1: x(ε)=101−εx(\varepsilon) = \frac{10}{1 - \varepsilon}.

  3. Evaluate the limit: lim⁡ε→0+101−ε=101−0=10\lim_{\varepsilon \to 0^+} \frac{10}{1 - \varepsilon} = \frac{10}{1 - 0} = 10.

Final answer

x(ε)=101−εx(\varepsilon) = \frac{10}{1 - \varepsilon}, with lim⁡ε→0+x(ε)=10\lim_{\varepsilon \to 0^+} x(\varepsilon) = 10

NEXT STEPS
Final Takeaways from Your Teacher

  • Review the formal piecewise definition of absolute value ∣A∣|A| and practice writing the domain conditions alongside each branch.

  • Verify your branch solutions by checking whether the value of xx actually satisfies the domain restriction of that branch.

  • Work through the flashcards below to solidify the concept of function branches.

Lock it in

Question

How do you define the positive branch when removing absolute value bars in ∣x−c∣=f(x)|x - c| = f(x)?

Answer

Set x−c=f(x)x - c = f(x) under the domain restriction x≥cx \ge c (where x−c≥0x - c \ge 0).

Similar question

01

For the equation ∣x−3∣=εx+5|x - 3| = \varepsilon x + 5, determine the exact solution x(ε)x(\varepsilon) on the positive branch (x≥3x \ge 3) for 0<ε<10 < \varepsilon < 1, and evaluate lim⁡ε→0+x(ε)\lim_{\varepsilon \to 0^+} x(\varepsilon).

Answer

x(ε)=81−εx(\varepsilon) = \frac{8}{1 - \varepsilon}, with lim⁡ε→0+x(ε)=8\lim_{\varepsilon \to 0^+} x(\varepsilon) = 8. Key steps: On x≥3x \ge 3, x−3=εx+5  ⟹  x(1−ε)=8  ⟹  x=81−εx - 3 = \varepsilon x + 5 \implies x(1 - \varepsilon) = 8 \implies x = \frac{8}{1 - \varepsilon}. Since 0<ε<10 < \varepsilon < 1, x>8≥3x > 8 \ge 3 holds, and lim⁡ε→0+81−ε=8\lim_{\varepsilon \to 0^+} \frac{8}{1 - \varepsilon} = 8.

02

For the same family ∣x−4∣=εx+6|x - 4| = \varepsilon x + 6, find the exact solution x(ε)x(\varepsilon) on the negative branch (x<4x < 4) for small positive ε\varepsilon, and evaluate lim⁡ε→0+x(ε)\lim_{\varepsilon \to 0^+} x(\varepsilon).

Answer

x(ε)=−21+εx(\varepsilon) = -\frac{2}{1 + \varepsilon}, with lim⁡ε→0+x(ε)=−2\lim_{\varepsilon \to 0^+} x(\varepsilon) = -2. Key steps: On x<4x < 4, −(x−4)=εx+6  ⟹  −x+4=εx+6  ⟹  x(1+ε)=−2  ⟹  x=−21+ε-(x - 4) = \varepsilon x + 6 \implies -x + 4 = \varepsilon x + 6 \implies x(1 + \varepsilon) = -2 \implies x = -\frac{2}{1 + \varepsilon}. For ε>0\varepsilon > 0, x<0<4x < 0 < 4 holds, and lim⁡ε→0+−21+ε=−2\lim_{\varepsilon \to 0^+} -\frac{2}{1 + \varepsilon} = -2.

03

For the equation ∣2x−6∣=εx+10|2x - 6| = \varepsilon x + 10 with small positive ε\varepsilon, find the solution x(ε)x(\varepsilon) on the positive branch (x≥3x \ge 3) and calculate lim⁡ε→0+x(ε)\lim_{\varepsilon \to 0^+} x(\varepsilon).

Answer

x(ε)=162−εx(\varepsilon) = \frac{16}{2 - \varepsilon}, with lim⁡ε→0+x(ε)=8\lim_{\varepsilon \to 0^+} x(\varepsilon) = 8. Key steps: On x≥3x \ge 3, 2x−6=εx+10  ⟹  x(2−ε)=16  ⟹  x(ε)=162−ε2x - 6 = \varepsilon x + 10 \implies x(2 - \varepsilon) = 16 \implies x(\varepsilon) = \frac{16}{2 - \varepsilon}. For small ε>0\varepsilon > 0, x≈8≥3x \approx 8 \ge 3, confirming validity. The limit as ε→0+\varepsilon \to 0^+ is 162=8\frac{16}{2} = 8.

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